What is the integral of 6x(x^2+9)^5 ?

The integral of 6x(x^2+9)^5 is 1/2 (x^2+9)^6+C

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integral of 6x(x^2+9)^5

\int 6 x (x^{2} + 9)^{5} d x

\frac{1}{2} (x^{2} + 9)^{6} + C

Solution steps

\int 6 x (x^{2} + 9)^{5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x (x^{2} + 9)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= 6 \cdot \int \frac{u ^{5}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \frac{1}{2} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 6 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} + 9

= 6 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 9 ) ^{6}}{6}

Simplify

6 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 9 ) ^{6}}{6} : \frac{1}{2} (x^{2} + 9)^{6}

= \frac{1}{2} (x^{2} + 9)^{6}

Add a constant to the solution

= \frac{1}{2} (x^{2} + 9)^{6} + C

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\int cos (π) t d t

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\frac{d}{d x} (sin (cos^{2} (x)))

n = 0 \sum \infty (\frac{4}{9})^{n}

What is the integral of 6x(x^2+9)^5 ?
The integral of 6x(x^2+9)^5 is 1/2 (x^2+9)^6+C